**3. The physics of colliding and exploding
objects**

The esoterica of colliding bodies can be difficult to grasp, but the basic principles need not be. This page explains the two basic principles and one or two of the corollaries most relevant to understanding JFK's motions as displayed in the Zapruder film after he was hit in the head from the rear.

**Universal principles
**There are two universal principles of the physics of
colliding bodies, first that momentum in any directions is conserved, and second
that total energy is conserved. The conservation of momentum is much easier to
apply than the conservation of energy, for it is the sum of potential and
kinetic energy that is conserved, and potential energy is often difficult or
impossible to observe directly.

Momentum is defied as the product of mass and velocity (

In other words, momentum is conserved in every collision, whereas kinetic energy is not always. (In fact, it is seldom conserved.) The special types of collisions that conserve kinetic energy are called

We can express these

*m _{1}v_{1} * +

where the prime refers to conditions after the collision. We can generalize this equation for any number of bodies by using the Greek Σ to designate the sum over all the bodies:

Σ * m _{i}v*

where i = 1, 2, 3, etc. For the usual x, y, and z axes, we can write:

Σ * p*_{xi} = Σ
* p _{xi}'*

Σ * p*_{yi} = Σ
* p _{yi}'*

Σ * p*_{zi} = Σ
* p _{zi}'*

(But the axes don't have to be perpendicular in order that momentum be conserved. It is conserved in any direction and in all directions simultaneously.)

For the conservation of energy, we can write:

*E*_{before} = * E*_{after}

which is the same as

*KE*_{before} + * PE*_{before} =
* KE*_{after} + * PE*_{after}

The notation with primes is simpler:

*KE* + * PE* = * KE'* + * PE'*

For a collection of particles, this becomes

Σ *KE*_{i} + Σ
* PE*_{i} = Σ *KE _{i}'* +
Σ

or

Σ * m _{i}v_{i}*

**Examples of conservation of momentum**

Let us first consider some qualitative implications of the
conservation of momentum. The simplest system of colliding particles is two
particles. Imagine one at rest and the other approaching from the left (with a
positive x-velocity). Let us say that the particles collide head-on, and that
the second particle moves off straight to the right. How will the first particle
be moving? The first thing we can say is that it will not be moving in the y or
z direction, because the first particle had no momentum in either of those
directions and the second particle also did not (after the collision).
Conversely, if the second particle moved off in the y direction (say upward),
the first particle would have to take on a velocity in the negative y direction
(downward) in order to make the net y-momentum after the collision sum to zero
(its value before the collision).

Let us now consider an actual example. Suppose the masses * m _{1}
*
and

*m _{1}v_{1x} * +

*v _{2x}'* = (

= (2x2 + 0 - 2x1.5) /1 = 1 fps

What will be the y-velocity of the second body? Do the same thing, but with y instead of x:

*m _{1}v_{1y} * +

*v _{2y}'* = (

= (2x0 + 0 - 2x0) /1 = 0 fps

But we knew that ahead of time because of the reasoning given above.

Now consider what happened to the kinetic energy in this
collision. We won't worry about the units of energy here, because they are too
tricky in the English system. We will just determine whether the kinetic energy
was conserved.

The initial * KE* was (2x2^{2})/2 + 0 = 4 "energy
units." The final * KE* was (2x1.5^{2})/2 + (1x1^{2})/2 = 2.75
energy units. Thus, about 30% of the kinetic energy was lost, making this an
inelastic collision. Where did that 30% of the * KE* go? The conventional answer is
that is was degraded into heat, but we shall see below that it can sometimes
wind up in other forms.

**Transformation of energy in extremely inelastic collisions**

Of interest of us is the special case of a fast-moving light
object striking a much heavier object and sticking to it. (This resembles the
bullet hitting JFK's head.) Imagine, for example, a 1-lb projectile moving at
2000 fps colliding with a 100-lb stationary object and sticking to it. We
address two questions: What is the final velocity of the pair, and what happens
to the kinetic energy in the process?

To find the final velocity of the pair, we use the same
equation for conservation of momentum that we used above, but with the same
final velocity for both objects, which we will call v_{1}'.

*m _{1}v_{1} * +

*v _{1}'* = (

= (1x2000 + 100x0)/(1 + 100) = 19.8 fps

That's the straightforward part. The kinetic energy part
is much more interesting. The initial * KE* is 1x2000^{2}/2 + 0 = 2,000,000
"energy units." The final * KE* is 101x19.8^{2}/2 = 19,798 units,
or only 1% of the original. In other words, 99% of the * KE* has disappeared in
this highly inelastic collision. Where did that 99% go? It disappeared into
deformation and something like heat, warming the deformed and fused bodies after they collided.

This very
significant aspect of inelastic collisions is seldom appreciated, but it is
vital to understanding the JFK assassination. It means that when a bullet hit
the head, the great majority of the initial *KE* will disappear into the
head, where it will temporarily be kept in reserve. If the head does not
explode, the energy will be degraded into heat and never seen or heard from
again. But if the head explodes, the *KE* can be released to work again,
this time in the form of explosive energy of fragments and of the head and body
in recoil to those fragments. The *KE* available this way is much greater
than any *KE* given to the body by the direct impact of the bullet. That is
why JFK's body can potentially recoil in any direction if his head explodes.
Think of the explosion as a way to release all that pent-up energy from the hit.

Ahead to 4--Wound Ballistics and Physics

Back to 2--Movements in the Z-film

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